Problem: The following equation is true for all real values of $x$ for which the expression on the left is defined, and $A$ is a polynomial expression. $\dfrac{A}{3x}\cdot\dfrac{3x+3}{x^2-2x-3}=1$ What is $A$ ? $A=$
Solution: The left side of the equation is a product of two rational expressions, while the right side is simply $1$. This means that the numerator and the denominator of the resulting product on the left side should cancel out completely. In order to solve for $A$, let's multiply the expressions and simplify as much as we can. We start with factoring the numerators and the denominators of the two expressions. [Why are we doing this?] The denominator of the first expression $3x$ cannot be factored any further. The numerator of the second expression $3x+3$ can be factored as $3(x+1)$ by factoring out a $3$. The denominator of the second expression $x^2-2x-3$ can be factored as $(x+1)(x-3)$ using the sum-product pattern. Now the product looks as follows: $\dfrac{A}{3x}\cdot\dfrac{3(x+1)}{(x+1)(x-3)}$ To find the product of two rational expressions, we multiply across, then simplify: [What's that?] $\phantom{=}\dfrac{A}{3x}\cdot\dfrac{3(x+1)}{(x+1)(x-3)}$ $\begin{aligned}&=\dfrac{A\cdot 3(x+1)}{3x\cdot (x+1)(x-3)}&\text{Multiply across.}\\\\\\ &=\dfrac{A\cancel{{(3)}}\cancel{{(x+1)}}}{\cancel{{3}}(x)\cancel{{(x+1)}}(x-3)}&\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{A}{x(x-3)}\end{aligned}$ After this simplification, our equation now looks like: $\dfrac{A}{x(x-3)}=1$ We can conclude that the numerator and the denominator of the expression on the left must be equal. [Why?] In other words, $A=x(x-3)$, which is equivalent to $x^2-3x$.